How do you multiply and simplify $(\frac { 2x y ^ { 2} \cdot - 2x ^ { 0} y ^ { 2} } { - 2x ^ { - 1} y ^ { 0} } ) ^ { 3}$ ?

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Answer 1 · 2018-06-26T03:35:32

The answer is $8x^6y^12$ .

$((2xy^2*(-2x^0 y^2))/(-2x^-1 y^0))^3$

First, remember that anything raised to the 0th power is equal to $1$ . So start by canceling out all the terms that are raised to $0$ .

$((2xy^2 * (-2(1)y^2))/(-2x^-1(1)))^3$

$((2xy^2 * (-2y^2))/(-2x^-1))^3$

Multiply the numerator by adding the exponent of like terms. Don't forget the negative sign!

$((-4xy^4)/(-2x^-1))^3$

$((2xy^4)/(x^-1))^3$

Remember that anything raised to the $-1$ is basically a reciprocal, so:

$x^-1 = 1/x$

Therefore:

$((2xy^4)/(1/x))^3$

which is

$(2xy^4 * x)^3$

$(2x^2y^4)^3$

Distribute the exponent

$2^3 * x^(2 * 3) * y^ (4*3)$

$= 8x^6y^12$

How do you multiply and simplify (\frac { 2x y ^ { 2} \cdot - 2x ^ { 0} y ^ { 2} } { - 2x ^ { - 1} y ^ { 0} } ) ^ { 3}(\frac { 2x y ^ { 2} \cdot - 2x ^ { 0} y ^ { 2} } { - 2x ^ { - 1} y ^ { 0} } ) ^ { 3}?